Timeline for What is a square root of a line bundle?
Current License: CC BY-SA 2.5
7 events
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| Nov 4, 2010 at 12:56 | comment | added | Kevin H. Lin | @r0b0t: Note that $c_1$ is additive w.r.t. tensor product. Note that $c_1$ is defined for $C^\infty$ complex vector bundles. Take anything with odd $c_1$. | |
| Nov 3, 2010 at 19:12 | comment | added | Andy Putman | No. Let $X$ be a complex manifold and $L$ be a holomorphic line bundle on $X$ that does not have a holomorphic square root. We then forget the complex structure, so $X$ is just a smooth manifold and $L$ is a smooth complex line bundle on $X$. By what I just said, $L$ (as a smooth line bundle) does not have a smooth square root. | |
| Nov 3, 2010 at 18:18 | comment | added | Vít Tuček | What I meant was to treat everything in the smooth category. I.e. does every line bundle on a smooth (orientable?) manifold have a square root? | |
| Nov 3, 2010 at 18:01 | comment | added | Andy Putman | If $L$ is a line bundle such that $L \otimes L$ is holomorphic, then $L$ is holomorphic, so a counterexample in the holomorphic category also gives a counterexample in the smooth category. | |
| Nov 3, 2010 at 17:56 | comment | added | Vít Tuček | And what about smooth line bundles? | |
| Nov 3, 2010 at 17:06 | comment | added | Andy Putman | It is not true that all holomorphic line bundles on complex manifolds have square roots. See Polizzi's answer. | |
| Nov 3, 2010 at 16:30 | history | answered | Vít Tuček | CC BY-SA 2.5 |